POJ #1018 Communication System

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简单dp题，0-1背包模型。定义d[i][j]表示考虑完前i个设备，最大bandwidth为j时的最小费用。bandwidth的范围题目中没说，但貌似不大（500以内）

#define UVa  "poj-1018.cpp"//Communication Systemchar fileIn[30] = UVa, fileOut[30] = UVa;#include<cstring>#include<cstdio>#include<algorithm>using namespace std;//Global Variables. Reset upon Each Case!const int maxn = 100 + 5, maxm = 500;int kase, n;int B[maxn][maxn], P[maxn][maxn], mi[maxn], d[maxm][2];/////void update(int& targ, int src) {    if(targ == -1 || targ > src) targ = src;}int main() {    scanf("%d", &kase);    while(kase --) {        scanf("%d", &n);        for(int i = 0; i < n;  i ++) {            scanf("%d", &mi[i]);            for(int j = 0; j < mi[i]; j ++) scanf("%d%d", &B[i][j], &P[i][j]);        }        int t = 0;        for(int i = 0; i < maxm; i ++) d[i][t] = -1;        for(int u = 0; u < n; u ++) {            for(int i = 0; i < maxm; i ++) d[i][t^1] = -1;            for(int i = 0; i < mi[u]; i ++) {                if(u == 0) update(d[B[u][i]][t^1], P[u][i]);                else for(int b = 0; b < maxm; b ++) {                    if(d[b][t] != -1)                        update(d[min(b, B[u][i])][t^1], d[b][t] + P[u][i]);                }            }            t ^= 1;        }        float ans = 0;        for(int i = 0; i < maxm; i ++) if(d[i][t] != -1) ans = max(ans, (float)i/(float)d[i][t]);        printf("%.3f\n", ans);    }    return 0;}

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